Question

First terms of two A.P.s are 2 and 5 and their common differences
are 3 and -2 respectively. If the difference of the seems of first'n'
terms of these two A.P.s is 195. Find the number of terms..
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Answer 1

Answer:

hii mate your answer is below:

let the two AP 's to be

AP•••••••••

ap..........

the first of the two AP's

A-2

a-5

The C.D.are

D=3

d= -2

According to the question

the difference of the sums of the n terms of the two AP's

=n/2(2A+(n-1)D) - n/2(2a+(n-1)d)=195

n/2 (2×2+(n-1)3) -(2×5+( n-1)(-2)=195

for calculation look at the pic

Step-by-step explanation:

hope it help if you like the answer please mark me as brainliest and follow too

Answer 2

Answer:

n = 10

Step-by-step explanation:

a =2

d =3

AP=2,5,8,11.......

sn=n/2(2a+(n-1) d)

n/2(2(2)+(n_1)3)

n/2(3n+1)..........(1)

a=5

d=-2

AP=5,3,1,-1,-3

sb=n/2(2a(n-1) d)

n/2(2(5)+(n-1) 2

n/2(10-2n+2)

n/2(12-2n)............(2)

according to the question

(n/2(3n+1))-(n/2(12-2n))=195

3n2/2 n n/2 - 12n/2 + 2n2/2 =195

5n2/2 - 11 n/2=195

n/2(5n-11)=195

5n2-11n=390

5n2-50n+39n-390=0

5n(n-10)+39(n-10)=0

(5n+39)(n-10) =10

n-10=0

n=10

5n+39

= -39/5

negative number cannot be AP or a fraction cannot be an AP

so

n = 10